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$$\lim_{x\to \infty} {\frac{x}{x^2+1} +\frac{x}{x^2+2} + ... + \frac{x}{x^2+x} }$$

It seems obvious that the result is zero for each term but in order to break the limit into its individual parts we must know that every term's limit exists .

Candamir
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Χριστίνα Κ.
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    Is x a natural number? This type of questions are usualy solved by the sandwich theorem. (I think it goes to 1 not to 0 by the way) – Yanko Jun 16 '17 at 14:28
  • this is not possible – Dr. Sonnhard Graubner Jun 16 '17 at 14:30
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    Hint for solving the question in a different way: $\frac{1}{n}>\frac{n}{n^2+k}>\frac{n}{n^2+n}$ for $k<n$ + sandwich theorem. – Yanko Jun 16 '17 at 14:33
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    We know a limit distributes over addition in the case that the number of terms you are taking the limit of stays constant. In your case, as $x$ gets larger, you get more and more terms in your expression, so we can't apply the previous idea here. – layman Jun 16 '17 at 14:33
  • Incidentally, at first glance, I was thinking of the technique that actually solves $\lim_{x\to\infty} \left(\frac{x}{x^2+1^2} + \frac{x}{x^2+2^2} + \cdots + \frac{x}{x^2 + x^2} \right) = \frac{\pi}{4}$. – Daniel Schepler Jun 16 '17 at 20:10

4 Answers4

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No, you can't do this. Here's a simpler example:

$$\lim_{n\to\infty} \frac1n + \frac1n + \cdots + \frac1n$$

where there are $n$ terms in the sum. If we add up the terms, then of course we have $\lim_{n\to\infty} 1 = 1$, even though each term is going to $0$.

The key is that the number of terms grows as $n$ tends to infinity. If there were a fixed number of terms, then you could take the sum of the limits of the terms.

Théophile
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  • A great example. Yes, in isolation, each term tends to $0$ as $n \to \infty$. But when you add more and more of them up, even though they are each getting smaller at each successive $n$, you are still adding more terms at each successive $n$, and the clear result (after taking the limit) is that, in this example, the act of adding more terms clearly overpowers the diminishing value of each term so that the sum, rather than diminishing itself, tends to a nonzero value. – layman Jun 16 '17 at 14:37
  • For more examples to totally demolish the idea that some people have that one can play with infinity to get around the problem of summing infinitely many terms, see this post. – user21820 Jun 17 '17 at 07:38
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Even if we speak about sequences, it is not possible to treat each term separately. The very simple example is the sequence $$\left(1+\frac{1}{n}\right)^n=\underbrace{\left(1+\frac{1}{n}\right)\cdot\left(1+\frac{1}{n}\right)\cdots\left(1+\frac{1}{n}\right)}_{n\text{ factors}}.$$ Each factor tends (separately) to $1$. Is it true that the limit is also $1$? No, this is $\text{e}$.

szw1710
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Note that the series of interest can be written as

$$\frac{x}{x^2+1}+\frac{x}{x^2+2}+\cdots +\frac{x}{x^2+x}=\sum_{k=1}^x \frac{x}{x^2+k}$$

The summand clearly satisfies the bounds

$$\frac{x}{x^2+x}\le\frac{x}{x^2+k}\le \frac{x}{x^2+1}$$

Hence we can assert that

$$\frac{x^2}{x^2+x}\le \sum_{k=1}^x\frac{x}{x^2+k}\le \frac{x^2}{x^2+1}$$

whereby application of the squeeze theorem yields the coveted limit

$$\lim_{x\to \infty}\left(\frac{x}{x^2+1}+\frac{x}{x^2+2}+\cdots +\frac{x}{x^2+x}\right)=1$$

Mark Viola
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Note that when $x\to \infty:$ $$\frac{x}{x^2+k} \sim \frac{1}{x}, \ \ \ 1\le k \le x$$ Hence: $$L \sim \lim_{x\to\infty} x \cdot \frac{1}{x} = 1.$$

farruhota
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